Cohomology of fibrations over the circle: how to compute the ring structure? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T02:34:13Z http://mathoverflow.net/feeds/question/11364 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/11364/cohomology-of-fibrations-over-the-circle-how-to-compute-the-ring-structure Cohomology of fibrations over the circle: how to compute the ring structure? algori 2010-01-10T20:28:40Z 2011-02-16T11:02:44Z <p>This question is inspired by <a href="http://mathoverflow.net/questions/4361/cohomology-of-fibrations-over-the-circle" rel="nofollow">http://mathoverflow.net/questions/4361/cohomology-of-fibrations-over-the-circle</a> Moreover, it can be considered a subquestion of the above, but somehow it seems to me that some of the more interesting points were not addressed there. So I decided to ask a bit more specific question to emphasize some of those points, but I would not mind at all if someone merges this question with the above.</p> <p>Let $E\to S^1$ be a fiber bundle with fiber $F$, and assume we know $H^{\bullet}(F,\mathbf{Q})$ as a ring and the monodromy action on it. Notice that since the base is a circle, the Leray spectral sequence degenerates in the second term for dimension reasons. So we have an exact sequence $0\to I\to H^{\bullet}(E,\mathbf{Q})\to Q\to 0$ where $I$ is the kernel and $Q$ is the image of $H^*(E,\mathbf{Q})\to H^{\bullet}(F,\mathbf{Q})$. In this sequence we know $Q,I$ and the action of $Q$ on $I$ from the Leray spectral sequence.</p> <ol> <li><p>Does this suffice to determine the rational cohomology of $E$ as a ring, up to isomorphism? My guess is that probably not, but I can't find a counter-example off hand.</p></li> <li><p>If not, would the higher Massey products on $H^*(F,\mathbf{Q})$ allow one to compute the cup product on $H^{\bullet}(E,\mathbf{Q})$?</p></li> <li><p>If not, would a rational homotopy model $A$ of the fiber suffice, together with an automorphism $A\to A$ that covers up to homotopy the "monodromy" automorphism of the differential forms on $F$? My guess is that probably yes, but notice that computing models of fibrations with non-simply connected bases can be tricky in general: take for example the space $X$ of all (ordered) couples of distinct points of the real projective plane and the projection on the first factor: the fiber is a M\"obius band, which contracts to the circle and the monodromy action changes the sign of the generator in degree 1; but we have $H^i(X,\mathbf{Q})=\mathbf{Q}$ if $i=0,3$ and zero otherwise.</p></li> </ol> http://mathoverflow.net/questions/11364/cohomology-of-fibrations-over-the-circle-how-to-compute-the-ring-structure/11810#11810 Answer by Ryan Budney for Cohomology of fibrations over the circle: how to compute the ring structure? Ryan Budney 2010-01-15T01:25:29Z 2010-01-15T01:25:29Z <p>The answer to (1) is no. The easiest example I've been able to think of is the case of a bundle over $S^1$ with fibre a compact surface (genus $\geq 2$). For your monodromy, choose a non-trivial element of the Torelli group (a diffeo that acts trivially on homology). The cup-product structure for the cohomology of this 3-manifold depends non-trivially on which element of the Torelli group you choose. This is very standard stuff in mapping class groups as it's closely related to something called the Johnson filtration. See for example:</p> <p>Johnson's homomorphisms of subgroups of the mapping class group, the Magnus expansion and Massey higher products of mapping tori. </p> <p>Teruaki Kitano. Topology and its applications. 69 (1996)</p> <p>Kitano answers your question (2) in this case. It seems likely that this argument is general, at least for bundles over $S^1$. </p> http://mathoverflow.net/questions/11364/cohomology-of-fibrations-over-the-circle-how-to-compute-the-ring-structure/11826#11826 Answer by Tom Church for Cohomology of fibrations over the circle: how to compute the ring structure? Tom Church 2010-01-15T05:43:16Z 2010-01-15T05:43:16Z <p>This is a continuation of Ryan's answer above, but it has become too large for a comment. I wanted to work out the details of Ryan's example explicitly, so that we can see explicitly where your conditions fail to determine the cohomology; perhaps this can help you to pin down precisely what conditions you want. It doesn't seem that we actually need Kitano here, just Johnson's classic results.</p> <p>Let $S_g\to M^3\to S^1$ be a mapping torus of an element of the Torelli group, i.e. a diffeomorphism $S_g\to S_g$ acting trivially on homology. Such a bundle admits cohomology classes satisfying the <a href="http://en.wikipedia.org/wiki/Leray-Hirsch%5Ftheorem" rel="nofollow">Leray-Hirsch condition</a> [this is a fun exercise], implying that as $H^{\ast}(S^1)$-modules, $H^\ast(M^3) = H^\ast(S_g) \otimes H^\ast(S^1)$. Thus the following do not depend on the monodromy:</p> <ul> <li>$Q = H^\ast(S_g)$,</li> <li>$I$, which is $Q$ with grading shifted by 1 (if $H^\ast(S^1) = \mathbb{Z}[t]/t^2$, this is $tQ$)</li> <li>the action of $Q$ on $I$ (just the action of $Q$ on $tQ$),</li> <li>and the Massey products on $Q = H^\ast(S_g)$ [although perhaps I misunderstand what you mean here].</li> </ul> <p>However, Johnson's work implies that your 3-manifold has the same cohomology ring as the product $S_g \times S^1$ iff the monodromy lies in the kernel of a certain homomorphism called the Johnson homomorphism; in particular, the ring $H^\ast(E)$ depends on the monodromy. It seems this shows that the answers to 1) and 2) are both "No".</p> <p>Now we can compare this with your conditions to see exactly what information we're missing; it turns out to be exactly the "Johnson homomorphism". The exact sequence above $0\to I\to H^\ast(E)\to Q\to 0$ has a splitting as abelian groups $H^\ast(E) = Q\oplus tQ$ coming from the Leray-Hirsch theorem as above. The only information we don't know automatically is the cup product on $Q$ in this splitting with itself. We know when we project back to the $Q$ factor we recover the cup product there, which means that the missing information is the projection onto the $tQ$ factor. Letting e.g. $Q(1)$ denote the degree 1 part, the cup product is a map $Q(1) \wedge Q(1) \to H^2(E)$. Projecting onto the $tQ$ factor, we have $Q(1) \wedge Q(1) \to tQ(2)$. But both $Q(1)$ and $tQ(2)$ are isomorphic to $H^1(S_g)$, so this projection of cup product is a map $\bigwedge^2 H^1(S_g) \to H^1(S_g)$. This exactly encodes the data that is not determined by your conditions; Johnson's beautiful result is that this map is exactly the Johnson homomorphism, originally defined from the algebraic properties of the monodromy. In particular he showed that this missing data could be zero or nonzero, and in fact can be anything in the subspace $\bigwedge^3 H^1&lt;\textrm{Hom}(\bigwedge^2 H^1,H^1)$.</p> <p>This was first laid out in Johnson's survey "A survey of the Torelli group" (<a href="http://www.ams.org/mathscinet-getitem?mr=718141" rel="nofollow">MR0718141</a>), and the details are worked out carefully in Hain, "Torelli groups and geometry of moduli spaces of curves" (<a href="http://www.ams.org/mathscinet-getitem?mr=1397061" rel="nofollow">MR1397061</a>). What Kitano is doing is different, or rather a generalization of this: showing that just as the cup product on $H^\ast(E)$ detects the Johnson homomorphism, the higher Massey products on $H^\ast(E)$ detect "higher Johnson homomorphisms" measuring deeper algebraic invariants. (If any of this is useful, please consider it a partial repayment for your beautiful summary of Hodge theory in <a href="http://mathoverflow.net/questions/11393/why-the-similarity-between-hodge-theory-for-compact-riemannian-and-complex-manifo/11407#11407" rel="nofollow">this answer</a>.)</p> http://mathoverflow.net/questions/11364/cohomology-of-fibrations-over-the-circle-how-to-compute-the-ring-structure/55609#55609 Answer by Diarmuid Crowley for Cohomology of fibrations over the circle: how to compute the ring structure? Diarmuid Crowley 2011-02-16T11:02:44Z 2011-02-16T11:02:44Z <p>Here's a high dimensional "no" to 1) and 2). Let $M = S^2 \times S^3$. There is an orientation preserving diffeomorphism of $f \colon M \cong M$ which induces the identity on all integral homology groups. However the map on $\pi_3(M) \cong \mathbb{Z} \oplus \mathbb{Z}$ is non-trivial - it is "triangluar" with the Hopf map in a corner and ones on the diagonal. If you take the mapping torus of this diffeomorphism, $T_f$, it is an oriented $6$-manifold: however there is a class $z \in H^2(T_f; \mathbb{Z})$ such that $z^3 \in H^6(T_f, \mathbb{Z})$ is a non-zero multiple of the fundamental class of $T_f$.</p>